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Question Number 6055 by 1771727373 last updated on 11/Jun/16

$$ln\left(x\right)+x=a$$$$x=?$$

Answered by Yozzii last updated on 11/Jun/16

$$lnx=a-x\Rightarrow x=e^{a-x}\Rightarrow {xe}^x=e^a$$$$\Rightarrow W\left({xe}^x\right)=W\left(e^a\right)$$$$x=W\left(e^a\right)$$$$W\left(z\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n-1}{n}^{n-2}}{(n-1)!}{z}^n(ref.Wolfram)$$$$\therefore W\left(e^a\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n-1}{n}^{n-2}{e}^{na}}{(n-1)!}$$$$x=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n-1}{n}^{n-2}{e}^{na}}{(n-1)!}$$

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